Lecture 9 - Boltzmanno number of 1-particle state...

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Unformatted text preview: Boltzmanno number of 1-particle state >>number of particles How many 1-particle states? Remember the sphere used to explain degeneracy? number of 1-particle states with an energy lower than = number of lattice points enclosed by the sphere in the positive 3 octant: ( ) 1 4 8 1 m = R3 = 8 3 6 h 2 3 a for thermal particles, 3 = = kT per particle 2 For the Bolztman condition to hold, 3 2 ( ) 1 m 3kT 8 = V 6 h 2 N 1 m 3kT V 8 N ? 1 6 h 2 m 3 2 3 2 ( ) ? 6 h which will occur for T =1 mkT 12 an d to express the energy of N particles, 1 ln Q ( N,V ,T ) Q ( N ,V , T ) E = - = - Q ( N ,V , T ) V V N - k 1 e N ! k E =- 1 q ( N,V ,T ) N! N N - k -1 1 N - k E =- e k ( - k ) e N N q ( N,V ,T ) ! k N! - k E = ke q ( N,V ,T ) k N E =N The average energy of N particles is N times the average energy of one particle e - k in terms of the probability , = k k where k = , - e j k # of available states ? N o we have to count more carefully we have to look at the particle's wavefunction Fermions: 2 particles cannot occupy the same quantum stateooccupation i1 i2 i3 k number 1,2 can only be 0 or 1 n P =- antisymmetric Bosons: 2 particles can occupy the same quantum stateooccupation number nk can have any value P1,2=+ symmetric i2 i2 i3 i4 i5 i6 There are many ways of counting distributions, 1 fluctuations of physical observables o X for large systems, fluctuations are negligible o all ensembles are equivalent choose Grand-Canonical Ensemble ( V ,T , ) = e N j - ENJ e - N ENj = energy states available for a system of N particles nk ( E j ) = molecules in k quantum state, in the E j state k = molecular quantum state { nk } defines the system at E j E j = k k and N = k n n e j -E NJ = e - i i i n k k a sum over states becomes the sum over each distribution { } ni ( V ,T , ) = N N e j - ENJ where we used = e - i i i - if we consider = e j -E NJ = e { n} n a sum over states becomes a sume over each distribution N =0 e n N - n i i i where n m m { } n k k =N = e N =0 - n i i i { n} ( - ( n11 +n2 2 +...+nk k +...) ( n1 +n2 +...+nk +...) ) = e N =0 { n} = N =0 { n} k ( e - k nk ) = n1=nmax n1=nmax 1 n L n =0 2 =0 k ( e - k nk ) where we use the fact that N can have any value (0 ) ( V,T, ) = nk =nmax k nk =0 ( e - k nk ) starting with ( V,T, ) = nk =nmax k nk =0 ( e - k nk ) we look at Fermions, w here nk =0,1 ( V, T , ) = = k nk =1 k nk =0 ( e - k nk ) - k e ( e ) +( ) 0 - k 1 ( V ,T , ) = k ( 1+ e - k ) Fermi-Dirac again, we start with ( V,T, ) = we now look at Bosons , where ( V, T , ) = nk = k nk =nmax k nk =0 ( e - k nk ) nk = 0,1,...N nk =0 ( e - k nk ) from our outstanding math background ... we recognize the series 1 x = 1 - x m =0 o m ( V ,T , ) = 1 k 1- e - k Bose-Einstein Combining Bose-Einstein with the Fermi-Dirac we obtain a gral. Eq. ( V ,T , ) = + + e - k ) - 1 ( 1- k for a Grand-canonical ensemble, N = kt and using = e kT ln ( V ,T , ) 1 kT d = e d kT ln ( V ,T , ) 1 ln ( V ,T , ) N = kt = kT + ln ( 1+ e - k ) e - k k N = = and N = + e - k ) k (1 nk k for each quantum state, the average number of particles nk = ( 1+ e - k ) - e - k e - k which is equivalent to j = e - k k we can now calculate all other thermo properties: k e - k E = N k = nk k = - k + ) k k ( 1 e and PV = kT ln ( V ,T , ) = kT k ln ( 1+ e - k ) - Keep in mind that even though we started with non-interacting particles, there wavefunctions are symmetrized othey are NOT INDEPENDENT to see if these eq. hold for Bolztmann statistics conditions Boltzmanno number of 1-particle state >>number of particles o nk 0, which will occur when 0 for 0 0 nk e - k = 1 e - k nk = e - k 1+ e - k ) ( nk N k e - k = k (Boltzmann-classical- limit) e - k and for the energy, E = lim E = k e - k 0 k k e - k k + - ( 1- e k ) k E N = e- = e- k k k k k e- = e - k k k k = finally, k k k (Boltzmann-classical limit) + + e - k ) - 1 =+ kT ( 1- pV = kt ln ( V ,T , ) = kT ln for pV = kT k 0 we can expand ln ( 1 x ) x + e - k = kt q(N,V ,T ) pV ln ( V ,T , ) = kt k k ln ( 1+ e - k ) - ( V ,T , ) = e q ( N,V ,T ) Examples: Photon gas, an electromagnetic field in thermal equilibrium with its container QM we know that H = i hi In=thisncase hi =harmonic oscillator i h To describe the state of the field, we need to know how many n are in each oscillator Photons are bosons o n=0,1,2,3..... Q ( N,V,T ) = - E = e n1 ,n2 ,...,n j e - ( n11 + n22 +...+ n j j ) first we want to show that Q ( N,V,T ) = n1 ,n2 ,...,n j e - ( n11 ) e - ( n22 ) L e - ( n j j ) = j n j =0 e- ( n j j ) Q ( N,V,T ) = ( e - ( 01 ) + e - ( 11 ) + - ( 21 ) ) e ) ( e - ( 0 + 3 = e - ( 01 + 02 ) + e - ( 01 +12 ) + e - ( 01 + 22 ) + e - ( 11 + 02 ) + e - ( 11 +12 ) + e - ( 11 + 22 ) + e - ( 21 + 02 ) + e - ( 21 +12 ) + e - ( 11 + 22 ) ) ( e - ( 0 + 3 ) e - ( 13 + e - ( 23 ) ) ( e - ( 0 ) 2 + e - ( 12 ) + e - ( 22 ) ) ) e - ( 13 + e - ( 23 ) ) = e - ( 01 + 02 + 0 3 ) + e - ( 01 + 02 +13 ) + e - ( 01 + 02 + 23 ) + e - ( 11 + 02 ) 1 2 3 1 2 +e - ( 01 +12 + 03 ) + e - ( 01 +12 +13 ) + e - ( 0 +1 + 2 ) + e - ( 0 + 2 + 0 3 ) 1 2 3 1 2 3 1 2 +e - ( 0 + 2 +1 ) + e - ( 0 + 2 + 2 ) + e - ( 1 + 0 + 03 ) + ... e - ( n11 + n22 + n33 ) n1 ,n2 ,...,n j 1 2 3 = e - ( 01 + 02 + 0 3 ) + e - ( 11 + 0 2 + 03 ) + e - ( 0 +1 + 0 ) 2 3 1 2 3 1 2 + e - ( 0 1 + 0 +1 ) + e - ( 2 + 0 + 0 ) + e - ( 0 + 2 + 0 3 ) + ... Q ( N,V,T ) = j e - ( nj j ) n j =0 = - ( nj j ) j 1- e 1 where we used xj = j 1 1- x The average number of photons in a state j... - E n j e - ( n11 + n22 +...) nje nj = = e - E { n} Q ln Q ( N,V ,T ) nj 1 nj = Q { n} n j e - ( n11 + n22 +...) - j ) n ( j = nj - ( j ) n ln Q ( N,V ,T ) = ( - 1) ln ( 1 - e - j ) ln Q ( N,V ,T ) - j ) ( 1 e + j e - j = 1 e - j - average occupation number is n j = = Planck - 1 distribution ...
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This note was uploaded on 05/29/2011 for the course CHM 6461 taught by Professor Bowers during the Spring '08 term at University of Florida.

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